A=84 and r=1/5
Since r, the common ratio squared is less than one, the sum will converge to a limit. Rule: if r^2<1 infinite series converges, otherwise it diverges.
Since the sum of any geometric sequence is:
s(n)=a(1-r^n)/(1-r)
And if r^2<1, (1-r^n) becomes 1-0=1 as n approaches infinity.
So whenever r^2<1 the sum of the infinite series is just:
s(n)=a/(1-r)
Since a=84 and r=1/5 this infinite series has a sum of:
s(n)=84/(1-1/5)
s(n)=84/(4/5)
s(n)=5(84)/4
s(n)=105
Answer:
blury
Step-by-step explanation:
Y = kx
so
k = y/x
<span>y is 20 when x is 4
k = 20/4 = 5
Answer
The </span><span>constant of variation for the relation = 5</span>
As we can assume the relationship is directly proportional, we have to build the proportion and do the cross product. So:
She needs 3 nights
Answer:
<h3>A,C,B 11,14,167#727272</h3>
